cp's OEIS Frontend

Viewed as a binary tree, this is (1); 5; 7,19; 11,23,29,65; ... Cf. A116623.

If we treat (2n+1) as a binary number with the nonzero bits numbered (highest bit first) from 0..k and the regular binary place value of each nonzero bit numbered from b(0) to b(k) then a(n) = 3^0 * b(0) + 3^1 * b(1) + .. + 3^k. For instance, if n=6 then 2n+1 = 13, which is equal to 8+4+1 or 1101 base(2); and a(n)=29 which is 8*1 + 4*3 + 1*9. - Joe Slater, Jan 23 2016

Related to the parity vectors of Terras and Collatz trajectories.

From Bob Selcoe, Sep 14 2019: (Start)

Let R_s be the reduced Collatz sequence starting with s and let R_s(i), i >= 0 be the i-th term in R_s. Then any term in R_s can be described as (3*s^i + k)/2^j, where j is the total number of halving steps from R_s(0) to R_s(i) i >= 1, and k is some term in A116641. k=1 when i=1; when i > 1, k is determined by the specific order of halving steps in R_s.

Ignoring duplicates, terms in A116641 > 1 can be generated by a series of subsequences:

1. Start with subsequence a(m) = 3+2^m, m >= 1; i.e., a(m) = {5,7,11,19,35,67,...}.

2. For fixed m, generate new subsequences b(n) = 3*a(m) + 2^(m+n), n >= 1; so:

m=1, a(1)=5, b(n) = 3*5 + {4,8,16,32,...} = {19,23,31,47,...};

m=2, a(2)=7, b(n) = 3*7 + {8,16,32,64,...} = {29,37,53,85,...};

m=3, a(3)=11, b(n) = 3*11 + {16,32,64,128,...} = {49,65,97,161,...}; etc.

3. Let 2^y be the summand used to find terms (t) in any previously-generated subsequence. (For instance, in m=2, b(3)=53: y=5 because t=53 = 3*7 + 32.) Continue generating new subsequences p(q) = 3*t + 2^(y+z) {z=1..inf} for all t. So in this example, from t=53 we get p(q) = 3*53 + {64,128,256,512,...} = {223,287,415,671,...}; from t=671 we get p(q) = 3*671 + {1024,2048,4096,...} = {3037,4061,6109,...}, etc. (End)

A subset of A116640 containing all terms A116640(m) such that m has binary weight of 2. This sequence is related to the Collatz and Terras trajectories; specifically those trajectories that include three odd numbers besides 1.

Create a binary tree starting with x. To follow 0 from the root, apply f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting with x, the string 010 {also known as f(g(f(x)))}, you would get (8x-2)/3. These expressions represent the reverse Collatz function and will provide numbers whose Collatz path may include x. These expressions will all be of the form (2^a*x-b)/3^c. This sequence concerns b. What makes b interesting is that if you draw the tree, each level of the tree will have the same sequence of values for b. The root of the tree x, can be written as (2^0*x-0)/3^0, which has the first value for b. Each subsequent level contains twice as many values of b.

This sequence is 0 followed by a permutation of A213539, and therefore consists of 0 plus the elements of A116640 multiplied by 2^k, where k >= 0. E.g., 1, 5, 7, 19 becomes 0, 2^0*1, 2^1*1, 2^0*5, 2^2*1, 2^0*7, 2^1*5, 2^0*19, ... - Joe Slater, Dec 19 2016

When this sequence is arranged as an irregular triangle the sum of each row a(2^k)...a(2^(k+1)-1) equals A081039(2^k). The cumulative sum from a(0) to a(2^k-1) equals A002697(k). - Joe Slater, Apr 12 2018

These numbers are of the form 3^k*2^{a_0} + 3^{k-1}*2^{a_1} + ... + 3^1*2^{a_{k-1}} + 3^0*2^{a_k} in which every power 3^i appears, 0 <= i <= k, and where a_i satisfies 0 <= a_0 < a_1 < ... < a_k.

These values are those of sequence A116640 in addition to any multiple of two of elements of this sequence. - Kenneth Vollmar, Jun 05 2013

Let R_s be the reduced Collatz sequence (cf. A259663) starting with s and let R_s(k), k >= 0 be the k-th term in R_s. Then R_(2n-1)(k) = (3^k*(2n-1) + T(n,k))/2^j, where j is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k). T(n,k) is defined here as the "residual summand".

The sequence without duplicates is a permutation of A116641.